**Question 17:**A 0.25 kg meterstick has a 2 kg mass suspended from the 80 cm mark and a 0.5 kg mass suspended from the 5 cm mark. At what mark should a pivot be placed to perfectly balance the meterstick?**Diagram Explanation:**The space below the question appears to be for solving the problem or for calculations. However, no graph or diagram is present in the image.**Answer 17:**___________To solve this problem, you need to apply the principle of moments (torque). The moment is calculated by multiplying the force (mass × gravity) by the distance from the pivot. Balance is achieved when the clockwise moments equal the counterclockwise moments.

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Answered: 17. A 0,25 kg meterstick has a 2kg mass…

17. A 0,25 kg meterstick has a 2kg mass suspended from the 80 cm mark and a 0.5 kg mass suspended from the 5 cm mark. At what mark should a pivot be placed to perfectly balance the meterstick?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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